Cambridge Prelim MX1 Textbook Marathon/Q&A (4 Viewers)

appleibeats · Aug 25, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

for integral -e to -2 of 1 -3x^2 / x - x^3 dx

I get using the result integral f'x / fx dx = log(fx)

the answer

log ( 6 / e^3 - e )

but in the answer it says log 6 / e^2 -1 - 1

I am not sure if I did it right or is there a way to simplify my answer??

rand_althor · Aug 25, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
for integral -e to -2 of 1 -3x^2 / x - x^3 dx

I get using the result integral f'x / fx dx = log(fx)

the answer

log ( 6 / e^3 - e )

but in the answer it says log 6 / e^2 -1 - 1

I am not sure if I did it right or is there a way to simplify my answer??

$\begin{align*}\ln{\frac{6}{e^3-e}}&=\ln\left [{\frac{1}{e} \times \frac{6}{e^2-1}} \right ] \\\textup{Since }\ln{ab}&=\ln{a}+\ln{b} \\&= \ln{\frac{6}{e^2-1}}+\ln{e^{-1}} \\&= \ln{\frac{6}{e^2-1}}-\ln{e} \\&= \ln{\frac{6}{e^2-1}}-1 \\\end{align*}$

appleibeats · Aug 28, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

For some reason I keep getting a negative in front of the 12.

My answer : - 12 - 4log4

The answer: 12 - 4log4

Question Find the area bewteen the are in the first quadrant enclosed between the two curves.

y = 4 /x and y = 1 + 4x - x^2

so

A = Integral 1 to 4 ( 1 + 4x - x^2 - 4/x) dx

= [ x + 2x - 1/3 x^3 - 4logx] 1 to 4

= ( - 28/3 - 4log4) - 8/3

= -12 - 4log4

Not sure where I have gone wrong??

InteGrand · Aug 28, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
For some reason I keep getting a negative in front of the 12.

My answer : - 12 - 4log4

The answer: 12 - 4log4

Question Find the area bewteen the are in the first quadrant enclosed between the two curves.

y = 4 /x and y = 1 + 4x - x^2

so

A = Integral 1 to 4 ( 1 + 4x - x^2 - 4/x) dx

= [ x + 2x - 1/3 x^3 - 4logx] 1 to 4

= ( - 28/3 - 4log4) - 8/3

= -12 - 4log4

Not sure where I have gone wrong??

$That emboldened $2x$ should be a $2x^2$, since integrating $4x$ gives $2x^2.$$

appleibeats · Aug 28, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

how is 2^n+1 - 2^n = 2^n

What indices law is that??

rand_althor · Aug 28, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
how is 2^n+1 - 2^n = 2^n

What indices law is that??

It isn't an indices law. You're factoring out 2^n like so:
$\begin{align*}2^{n+1}-2^{n}&=2^n\times2^1-2^n \\&=2^n(2-1) \\&=2^n\times 1 \\&=2^n\end{align*}$

appleibeats · Aug 28, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Why is the integral of e^kx = 1/k . e^kx

isnt it just e^kx . k

why is there a fraction ??

rand_althor · Aug 28, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$$For $f(x)=ax+b$, where $a$ and $b$ are arbitrary constants, integrating $e^{f(x)}$ gives you $\frac{1}{f'(x)}e^{f(x)}+C$, whereas differentiating $e^{f(x)}$ gives you $f'(x)e^{f(x)}$. So the integral of $e^{kx}$ is $\frac{1}{k}e^{kx}+C$, while the derivative of $e^{kx}$ is $ke^{kx}$. Perhaps you are confusing the two?$

InteGrand · Aug 28, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

rand_althor said:
$Integrating $e^{f(x)}$ gives you $\frac{1}{f'(x)}e^{f(x)}+C$

$That's not true for a general $f(x)$ of course (in fact, it is true if and only if $f(x)=ax+b$ for some constants $a$ and $b$ (i.e. $f$ is a linear or constant function)).$

rand_althor · Aug 28, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

InteGrand said:
$That's not true for a general $f(x)$ of course (in fact, it is true if and only if $f(x)=ax+b$ for some constants $a$ and $b$).$

Ah right my bad. Thanks for the correction.

appleibeats · Aug 28, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

It is known that f'(x) = e^x + 1/e and that f(-1) = -1 . Find f(0)

Integral e^x + 1/e dx

= e^x + loge + C

so

e^-1 + 1 + C = -1

C = - 1/e - 2

So I get

f(x) = e^x - 1/e - 1

However the answer has and x over the fraction. I can't see where I have gone wrong in my working?

ANSWER: f(x) = e^x - x/e - 1

InteGrand · Aug 28, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
It is known that f'(x) = e^x + 1/e and that f(-1) = -1 . Find f(0)

Integral e^x + 1/e dx

= e^x + loge + C

so

e^-1 + 1 + C = -1

C = - 1/e - 2

So I get

f(x) = e^x - 1/e - 1

However the answer has and x over the fraction. I can't see where I have gone wrong in my working?

ANSWER: f(x) = e^x - x/e - 1

$The integral of $\frac{1}{e}$ is $\frac{1}{e}x$, not $\ln e$ (since you're integrating a constant).$

appleibeats · Aug 28, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

for the equation y = -e^2-x + e^2 + 1

How do you work out the horizontal asymptote.

The answer is e^2 + 1

InteGrand · Aug 28, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
for the equation y = -e^2-x + e^2 + 1

How do you work out the horizontal asymptote.

The answer is e^2 + 1

$For horizontal asymptotes, observe the behaviour of $y$ as $x\to \infty$ and as $x\to -\infty$.$

appleibeats · Aug 28, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Not sure how to find teh primative of 1 /x^1/2 - xe^-x^2

InteGrand · Aug 28, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Not sure how to find teh primative of 1 /x^1/2 - xe^-x^2

Standard integration of a power of x for the first one, and reverse chain rule for the second.

appleibeats · Aug 28, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

how do you find primative of log (e^2kx)

InteGrand · Aug 28, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
how do you find primative of log (e^2kx)

$Assuming you mean $\ln \left(e^{2kx} \right)$, that is just equal to $2kx$, which is easy to find the primitive of.$

appleibeats · Aug 29, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

so differentiate xlogx

I get how d/dx (xlogx) = 1 + logx

Product rule

And so reversing this;

integral of ( 1 + logx) dx = xlogx + C

This next step i don't get

Integral of logx dx = xlogx - x + C

I don't get how to get the -x and where did the 1 go??

InteGrand · Aug 29, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
so differentiate xlogx

I get how d/dx (xlogx) = 1 + logx

Product rule

And so reversing this;

integral of ( 1 + logx) dx = xlogx + C

This next step i don't get

Integral of logx dx = xlogx - x + C

I don't get how to get the -x and where did the 1 go??

$We have$

$$\frac{\mathrm{d}}{\mathrm{d}x}\left(x\ln x \right)=\ln x + 1$. Rearranging,$

$$\ln x = \frac{\mathrm{d}}{\mathrm{d}x}\left(x\ln x \right) -1$.$

$Integrating both sides, $\int \ln x \text{ d}x = \int \left( \frac{\mathrm{d}}{\mathrm{d}x}\left(x\ln x \right) -1 \right)\text{ d}x$$

$$=\int \frac{\mathrm{d}}{\mathrm{d}x}\left(x\ln x \right) \text{ d}x-\int 1 \text{ d}x$ (since the integral of a difference is the difference of the integrals)$

$$= x\ln x - x + C$.$

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